A person pushes a 10.0 kg lawn mower at constant speed with a force of

nbroyle1

A person pushes a 10.0 kg lawn mower at constant speed with a force of 82.0 N directed along the handle, which is at an angle of θ = 45.0° to the horizontal (Fig. 4-45).

(b) Calculate the horizontal retarding force on the mower
(c) Calculate the normal force exerted vertically upward on the mower by the ground.
(d) Calculate the force the person must exert on the lawn mower to accelerate it from rest to 1.0 m/s in 2.0 seconds (assuming the same retarding force).

So I've already solved for the horizontal force by using trig and it was 57.98N. I also found the normal force by adding weight and the vertical component of the force. Now Im trying to find the force the person must exert on the lawn mower to accelerate it with the variables given.
Is this involving kinematics what should I focus on for this step???

Xisune

The normal force would be weight - the vertical component, since adding a force upwards would weaken the normal force.

As for the last question, find the acceleration using kinematics, then use f=ma

nbroyle1

I tried that and got a=1/2. so when I plugged that into f=ma I got f=10(1/2) but the answer is incorrect. am I forgetting something or is my calculations just wrong??

nbroyle1

yes but since the person is pushing on the lawn mower you add the vertical component of his push + mg and that gives you the normal force.

nbroyle1

I did the equation V=Vo+at so after I plugged in the information and solved for acceleration it was 1/2. If I plug that into f=ma I get 5 but that isn't right.

Xisune

Did the question specify that the force must be 45°? If not then you can just add that force to the retarding force.



If the force is 45 degrees above, then the vertical force is up, not down, thus you subtract.

PhanthomJay

Yes, correct

The force you have solved is the net force in the horizontal direction. You must subtract the retarding force to get the appled force in the horizontal direction, then find the applied force along the handle using trig.